Answer:
c. Fission and fusion are two processes that release very little amounts of energy.
Explanation:
This statement is false. In fact, both fission and fusion are processes which release very large amounts of energy. The statement can be rewritten as it is true as follows:
"Fission and fusion are two processes that release very large amounts of energy."
Fission occurs when a large nucleus break apart, splitting into smaller nuclei, while fusion occurs when two light nuclei combine together into a larger nucleus. In both cases, the mass of the reactants is larger than the mass of the final products, so some of the mass has been converted into energy, according to Einstein's equation:
where
E is the energy released
is the mass lost in the process
c is the speed of light
Since c is a very large number (
), we see that even a very small mass causes the released of a huge amount of energy, so both fission and fusion release large amounts of energy.
Answer:
1. A fossil that is widespread geographically but only occurs in one layer or a small number of layers of rock
2. Sediment deposited into water will spread in a horizontal and continuous sheet
3. Any undisturbed sequence of layered rocks has the oldest rock on the bottom and newest rock on the top
4. Study of fossils
5. Study of rock layers and the process that form them
Sorry if any of these are wrong
I will assume that big Joe is big Jim. The equation for the momentum is p=m*v, where m is the mass of the body and v is the velocity. Big Joe has a mass m=105 kg and speed v=5.2 m/s. When we input the numbers:
p=105*5.2=546 kg*(m/s).
So big Joe's momentum before the collision is p=546 kg*(m/s).
Answer:
right; left
Explanation:
The right hemisphere of the brain has a schema formation capacity, this can be applied to recognize faces. The left hemisphere is in charge of the memory and therefore to recall names.
I hope you find this information useful and interesting! Good luck!
Answer:
the volume decreases at the rate of 500cm³ in 1 min
Explanation:
given
v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min
PV=C
vΔp + pΔv = 0
differentiate with respect to time
v(Δp/t) + p(Δv/t) = 0
(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0
40000 + 80kPa(Δv/t) = 0
Δv/t = -40000/80
= -500cm³/min
the volume decreases at the rate of 500cm³ in 1 min
